# Elementary Number Theory Problems 4.2 Solution (David M. Burton's 7th Edition) - Q5

My Solution "Prove that the integer $53^{103} + 103^{53}$ is divisible by $39$, and that $111^{333}$ + $333^{111}$ is divisible by $7$."

## Background

All theorems, corollaries, and definitions listed in the book's order:

I will only use theorems or facts that are proved before this question. So you will not see that I quote theorems or facts from the later chapters.

## Question

Prove that the integer $53^{103} + 103^{53}$ is divisible by $39$, and that $111^{333}$ + $333^{111}$ is divisible by $7$.

## Solution

For $53^{103} + 103^{53}$:

We have $53^{2} \equiv 14^2 \equiv 1 \pmod {39}$, and $103^{2} \equiv 25^2 \equiv 1 \pmod {39}$.

Thus we know

$$53^{103} + 103^{53} \equiv (53^{2})^{51} \cdot 53 + (103^{2})^{26} \cdot 103 \equiv 53 + 103 \\ \equiv 156 \equiv 0 \pmod {39}$$

Therefore, $53^{103} + 103^{53}$ is divisible by $39$.

For $111^{333} + 333^{111}$:

We first have $111 \equiv -1 \pmod 7$. Then $111^{333} \equiv -1 ^{333} \equiv -1 \pmod 7$.

Consider $333^{111} \equiv (111 \times 3)^{111} \equiv -1^{111} \cdot 3^{111} \equiv -1 \cdot 3^{111} \pmod 7$.
Since $3^{3} \equiv -1 \pmod 7$, so $-1 \cdot 3^{111} \equiv -1 \cdot (-1)^{37} \equiv 1 \pmod 7$.

So, $111^{333} + 333^{111} \equiv -1 + 1 \equiv 0 \pmod 7$. Therefore, $111^{333}$ + $333^{111}$ is divisible by $7$.

Read More: All My Solutions for This Book

MathNumber TheorySolution